Question

an object of size 4cm is placed at a distance of 15 cm Infront of a concave mirror of focal length 10 cm. find the size position and nature of the image​

Answer 1

Solution

Given :-

• Focal length of concave mirror = 10 cm
• Size of object = 4 cm
• Distance of object from focal length = 15 cm

Find :-

• Image position and its nature

Explanation

Using mirror Formula

$\boxed{\tt{\red{\:\dfrac{1}{f}\:=\:\dfrac{1}{v}+\dfrac{1}{u}}}}$

Where,

• f = focal length
• v = distance of image from focal length
• u = distance of object from focal length

Keep all values in formula

==> 1/(-10) = 1/(-15) + 1/v

==> 1/v = -1/10 + 1/15

==> 1/v = (-15 + 10)/150

==> 1/v = (-5)/150

==> 1/v = -1/30

Or,

==> v = - 30 cm.

Distance of image is (-ve) sign , that means image be real .

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Some important

• concave mirror are called converging mirrors.
• focal length of concave mirror always be negative
• all distance will be positive , if we read distance R.H.S. from to concave mirror .
• all distance will be negative , if we read distance L.H.S. from to concave mirror .

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Answer 2

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Given :-

• Focal length of concave mirror = 10 cm

• Size of object = 4 cm

• Distance of object from focal length = 15 cm

To Find :-

• Image position and its nature.

Explanation :-

Using Mirror Formula :-

$\boxed{\tt{\pink{\:\dfrac{1}{f}\:=\:\dfrac{1}{v}+\dfrac{1}{u}}}}$

Where,

• f = - 10 cm
• h = - 15 cm

Keep all values in formula

• ➟ 1/(-10) = 1/(-15) + 1/v

• ➟ 1/v = -1/10 + 1/15

• ➟ 1/v = (-15 + 10)/150

• ➟ 1/v = (-5)/150

• ➟ 1/v = -1/30

Or,

• ➟ v = - 30 cm.

• Distance of image is (-ve) sign , that's means image be real .

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